Problem: The equation of a circle $C$ is $x^2+y^2+18x+12y+68 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Explanation: To find the equation in standard form, complete the square. $(x^2+18x) + (y^2+12y) = -68$ $(x^2+18x+81) + (y^2+12y+36) = -68 + 81 + 36$ $(x+9)^{2} + (y+6)^{2} = 49 = 7^2$ Thus, $(h, k) = (-9, -6)$ and $r = 7$.